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A 50 Hz, 11 kV, 3 phase alternator with earthed neutral having a reactance of 3 ohms per phase and is connected to a bus bar through a circuit breaker, if the distributed capacitance upto CB between the phase and neutral is 0.01 μ F.
What is the peak re striking voltage?
- a)18.36 kV
- b)17.96 kV
- c)15.96 kV
- d)12.65 kV
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A 50 Hz, 11 kV, 3 phase alternator with earthed neutral having a react...
Microfarad per phase, then determine the short circuit current available at the bus bar if a fault occurs at the CB.
To find the short circuit current, we need to first calculate the total impedance of the system. The reactance of the alternator per phase is given as 3 ohms, and the distributed capacitance per phase up to the CB is 0.01 microfarad.
The reactance of the distributed capacitance can be calculated as:
Xc = 1 / (2πfC) = 1 / (2π x 50 x 0.01 x 10^-6) = 318.3 ohms
The total impedance per phase is then given by:
Z = √(R^2 + Xl^2 + Xc^2) = √(3^2 + 318.3^2 + 3^2) = 318.9 ohms
Since we have a 3 phase system, the total impedance will be 3 times this value, which is 956.7 ohms.
The short circuit current can be calculated using Ohm's law:
I = V / Z
Where V is the voltage of the system, which is 11 kV or 11,000 volts.
I = 11,000 / 956.7 = 11.5 kA
Therefore, the short circuit current available at the bus bar is 11.5 kA.
To find the short circuit current, we need to first calculate the total impedance of the system. The reactance of the alternator per phase is given as 3 ohms, and the distributed capacitance per phase up to the CB is 0.01 microfarad.
The reactance of the distributed capacitance can be calculated as:
Xc = 1 / (2πfC) = 1 / (2π x 50 x 0.01 x 10^-6) = 318.3 ohms
The total impedance per phase is then given by:
Z = √(R^2 + Xl^2 + Xc^2) = √(3^2 + 318.3^2 + 3^2) = 318.9 ohms
Since we have a 3 phase system, the total impedance will be 3 times this value, which is 956.7 ohms.
The short circuit current can be calculated using Ohm's law:
I = V / Z
Where V is the voltage of the system, which is 11 kV or 11,000 volts.
I = 11,000 / 956.7 = 11.5 kA
Therefore, the short circuit current available at the bus bar is 11.5 kA.
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Community Answer
A 50 Hz, 11 kV, 3 phase alternator with earthed neutral having a react...
Max Restriking V=2x Max of(Vphase ) 2x sqrt(2) x11 / sqrt(3) = 2x1.414x6.35 =17.96
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A 50 Hz, 11 kV, 3 phase alternator with earthed neutral having a reactance of 3 ohms per phase and is connected to a bus bar through a circuit breaker, if the distributed capacitance upto CB between the phase and neutral is 0.01 μ F.What is the peak re striking voltage?a)18.36 kVb)17.96 kVc)15.96 kVd)12.65 kVCorrect answer is option 'B'. Can you explain this answer?
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